The brakes applied to a car produce an acceleration of $6\, m s ^{-2}$ in the opposite direction of motion. If the car takes $2\, s$ to stop after the application of the brakes, calculate the distance it travels during this time.
Given $a=-6 m s ^{-2}, t=2 s , v=0 m s ^{-1}$
$(i)$ Using the equation $v=u+a t,$ we have
$u=v-a t=0-(-6) \times 2$
$=12 m s ^{-1}$
$(ii)$ Now, using the expression $S=u t+1 / 2 a t^{2}$
we have $S=12 \times 2+\frac{1}{2} \times-6 \times(2)^{2}=12 m$
A body thrown in the vertically upward direction rises upto a height $'h^{\prime}$ and comes back to the position of its start.
Calculate :
$(a)$ the total distance travelled by the body and
$(b)$ the displacement of the body. Under what condition will the magnitude of the displacement be equal to the distance travelled by an object ?
The velocity-time graph of cars $A$ and $B$ which start from the same place and move along a straight road in the same direction is shown below
Calculate :
$(a)$ the acceleration of car $B$ between $2 \,s$ and $4\, s$.
$(b)$ the time at which both the cars have the same velocity.
$(c)$ the distance travelled by the two cars $A$ and $B$ in $8\, s$
$(d)$ Which of the two cars is ahead after $8\, s$ and by how much ?
A hiker rides $700\, m$ north, $300$ meast, $400 \,m$ north $600\, m$ west, $1200\, m$ south, $300\, m$ east and finally $100\, m$ north. Draw the path of motion of the biker. What distance did he cover ? What was his displacement ?
A cyclist goes once around a circular track of diameter $105$ metre in $5$ minutes. Calculate his speed.
What determines the direction of motion of an object velocity or acceleration ?